In Mathematics, hyperbolic functions are similar to trigonometric functions but are defined using the hyperbola rather than the circle. Just as the points (sin t, cost t) in trigonometry form a unit circle with radius, the points ( sinh t, cosh t) form the right half of the unit parabola.
Also, the derivatives of sin(t), and cos (t) in trigonometry are cos (t) and – sin(t) respectively, the derivatives of sinh(t), and cosh(t) in hyperbolic functions are cosh(t) + sinh(t).
Hyperbolic functions are shown up in the calculation of angles and distance in hyperbolic geometry. They are also shown up in the solutions of many linear differential equations, cubic equations, and Laplaces’ equations in cartesian coordinates.
Here, we will discuss hyperbolic trigonometric functions, hyperbolic trigonometric formulas, hyperbolic trigonometric identities, Properties of hyperbolic functions, derivatives of hyperbolic functions, integration of hyperbolic functions, etc. |
Properties of Hyperbolic Functions
The properties of hyperbolic functions are similar to the properties of trigonometric functions. Following are important properties of hyperbolic functions:
Hyperbolic Functions can also also be derived from the trigonometric functions with complex arguments.
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Sinh y = i sin(iy)
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Cosh y = cos (iy)
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Tanh y = -i tan(iy)
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Sech y = sec(iy)
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Cosech y = i cosec (iy)
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Coth y = i cot(iy)
Derivatives of Hyperbolic Functions
Following are the six derivatives of hyperbolic functions:
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[frac{d}{dy}] sinh y = cosh y
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[frac{d}{dy}] cosh y = sinh y
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[frac{d}{dy}] tanh y = 1- tanh² y = sech² y = [frac{1}{Cosh^{2}y}]
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[frac{d}{dy}] sech y = – sech y tanh y
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[frac{d}{dy}] csch y = – csch y coth y
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[frac{d}{dy}] coth y = 1- coth² y = -csch² y = [frac{1}{Sinh^{2}y}]
Integration of Hyperbolic Functions
As hyperbolic functions are defined in terms of eˣ and e⁻ˣ, we can easily derive rules for their integration. Following are all the six integration of hyperbolic functions:
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∫ coshy dy = sinh y + C
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∫ sinhy dy= cosh y + C
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∫ sech²y dy = tanh y + C
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∫ csch²y dy = – coth y + C
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∫ sech y tanh y dy = – sech y + C
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∫ csch y coth y dy = – csch y + C
Hyperbolic Function Graphs
Hyperbolic Function Graphs are given below:
Hyperbolic Sine Function (Sinh x)
Domain: R
Range: R
Hyperbolic Cosine Function (Cosh x)
Domain: R
Range: (1 + ∞)
Hyperbolic Tangent Function (Tanh x)
Domain: R
Range: ( -1 + 1)
Hyperbolic Cotangent x
Domain: (- ∞, 0) U (0 + ∞)
Range: (- ∞ – 1)(1 + ∞)
Hyperbolic Secant x
Domain: R
Range: ( 0, 1)
Hyperbolic Cosecant x
Domain: ( -∞ , 0) U (0 + ∞)
Range: ( -∞ , 0)(0 + -∞)
Hyperbolic Trigonometric Functions
Hyperbolic trigonometric functions define the idea of parametric equations for a unit circle (x = cost y and y = sin t) to the hyperbola’s parametric equation, which obtains the following two equations of hyperbola:
x = cosh (a) = [frac{e^{a} + e^{-a}}{2}] y = sinh (a) = [frac{e^{a} – e^{-a}}{2}]
A very important fact that should be considered is that the hyperbolic trigonometric functions consider the area as their argument (known as “ the hyperbolic angle”), as represented below:
Hyperbolic Trigonometric Formulas
The basic hyperbolic trigonometric formulas for sinh x and cosh x are defined by the exponential function eˣ and its inverse exponential function e⁻ˣ. Here e is the Euler’s constant. Let us understand the hyperbolic trigonometric formulas one by one.
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Hyperbolic Sine Function
The hyperbolic function f(x)= sinh x is defined as:
Sinh(x) = [frac{e^{a} – e^{-a}}{2}]
This function satisfies the condition of sinh (-x) = – sinh x and sinh 0 = 0
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Hyperbolic Cosine Function
The hyperbolic function f(x)= cosh x is defined as:
Cosh (x) = [frac{e^{a} + e^{-a}}{2}]
This function satisfies the condition of cosh (-x) = – cosh x and cosh 0 = 1
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Hyperbolic Tangent Function
The hyperbolic function f(x)= tanh x is defined as:
Tanh (x) = [frac{e^{a} – e^{-a}}{e^{a} + e^{-a}}]
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Hyperbolic Secant Function
The hyperbolic function f(x) = sech x is defined as:
Sech (x) = [frac{1}{cosh x}]
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Hyperbolic Cosecant Function
The hyperbolic function f(x) = csch x is defined as:
Csch (x) = [frac{1}{sinh x}]
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Hyperbolic Cotangent Function
The hyperbolic function f(x) = coth x is defined as:
Coth (x) = [frac{e^{a} + e^{-a}}{e^{a} – e^{-a}}]
Hyperbolic Trigonometric Identities
Hyperbolic trigonometric identities are similar to standard trigonometric identities and can be understood from the equations given below:
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Sinh y = – sinh(-y)
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Cosh y = cos(-y)
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Tanh y = -tanh(-y)
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Sech y = sech(-y)
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Coth y = -coth(-y)
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Csch y = -csch (-y)
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Sinh (α + λ) = sinh α cosh λ + cosh a sinh λ
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Sinh (α – λ) = sinh α cosh b – cosh sinh λ
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Cosh (α + λ) = cosh α cosh λ + cosh sinh λ
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cosh (α – λ) = cosh α cosh λ – cosh cosh λ
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Sinh2 y = [frac{-1+cosh(2y)}{2}]
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Sinh (2y) = 2 sinh2 y cosh2 y
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Cosh2 y = [frac{1+cosh(2y)}{2}]
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Cosh(2y) = cos2 y + sin2 y
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Cosh2 (y) – Sinh2 (y) = 1
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Tanh2 (y) + Sech2 (y) = 1
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Coth2 (y) – Csch2 (y) =1
Applications of Hyperbolic Functions
There numerous applications of the hyperbolic functions;
For instance, a company plans on building a suspension bridge which will stretch between the basketball arena and the baseball stadium, up until the other side of the railway line in a particular city. The center of this bridge will be suspended between the two concrete pillars that are 280 feet apart and have 80 feet height. The cable that is holding the bridge should be exactly 30 feet above the railway tracks and in the middle of the bridge, that is, it should sag exactly 50 feet.
Gottfried leibniz and Christian Huygens, in the year 1691, figured that any cable that hangs under the gravitational force should have the shape of the graph
y(x) = a cosh (x/a) + b
This shape came to be known as the catenary. The parameter a here represents the ratio of the cable tension to the cable density. The parameter here is used to represent a vertical shift if needed.
Two questions can arise given the situation here. Firstly, what values should the parameters a and b possess for the catenary to fit the constraints as given by the concrete pillars placement and the cable’s low point.
To find the values of a and b, we will have to look into two different questions. As we already know that y(0) = 30. This ensures that there is sufficient clearance present above the railway lines. We are also given the fact that y(140) = 80, as the cable is attached to an 80 foot tall pillar and 140 feet from the lowest point, that is the centre. Thus,
30= a cosh (0/a) + b
80 = a cosh (140/a) +b
As cosh(0)= , we can easily simplify the first equation as 30= a + b. While substituting this new derived equation into the second equation we get 80 = a cosh (140/a) + 30- a.
After we put these equations on a graph, that is , the function y= x cosh (140/x) + 30 -x , where the line y= 80 , and thus searching the point of intersection, we notice that 80 = a cosh (140/a) + 30 – a where a ^^ 203.82. Using this value together with 30= a + b , we will get b= -173.82.
Thus, we can model the height of the bridge with the equation
h(x) = 203.82 cosh (x/203,82)-173.82
Solved Example:
1. Simplify [frac{Sinh x}{Cosh y}]
Solution:
As we know,
Sinh(x) = [frac{e^{a} – e^{-a}}{2}] and Cosh (x) = [frac{e^{a} + e^{-a}}{2}]
Accordingly,
[frac{Sinh x}{Cosh x}] = [frac{frac{e^{a} – e^{-a}}{2}}{frac{e^{a} + e^{-a}}{2}}]
[frac{Sinh x}{Cosh x}] = [frac{e^{a} – e^{-a}}{e^{a} + e^{-a}}]
Therefore, [frac{Sinh x}{Cosh x}] = [frac{e^{a} – e^{-a}}{e^{a} + e^{-a}}]
2. Express eˣ and e⁻ˣ as a Function of sinh x and cosh x.
Solution:
Adding two equations of cosh x and sinh x gives:
Cosh x + Sinh x = [frac{e^{x} + e^{-x}}{2}] + [frac{e^{x} – e^{-x}}{2}] (2)
Cosh x + Sinh x = eˣ
Subtracting two equations of cosh x and sinh x gives:
Cosh x – Sinh x = [frac{e^{x} + e^{-x}}{2}] – [frac{e^{x} – e^{-x}}{2}]
Cosh x – Sinh x = e⁻ˣ (2)
Combining equation (1) and equation (2), we get
e±x = cosh x ± sinh x
Which is equivalent to Euler’s formula of hyperbolic function.